1. **Problem 2a:** Determine whether $\neg q \wedge (p \to q) \to p$ is a tautology or contradiction using logical equivalences. 2. Recall that $p \to q$ is logically equivalent to $\neg p \vee q$. 3. Rewrite the expression: $$ (\neg q \wedge (p \to q)) \to p = (\neg q \wedge (\neg p \vee q)) \to p $$ 4. Use the implication equivalence $a \to b \equiv \neg a \vee b$: $$ \neg (\neg q \wedge (\neg p \vee q)) \vee p $$ 5. Apply De Morgan's law to $\neg (\neg q \wedge (\neg p \vee q))$: $$ \neg (\neg q) \vee \neg (\neg p \vee q) \vee p = q \vee (p \wedge \neg q) \vee p $$ 6. Simplify $q \vee (p \wedge \neg q) \vee p$: Since $p \vee (p \wedge \neg q) = p$, the expression becomes: $$ q \vee p \vee p = p \vee q $$ 7. The entire expression simplifies to: $$ p \vee q $$ 8. Since $p \vee q$ is not always true or always false, the original formula is **neither a tautology nor a contradiction**. --- 1. **Problem 2b:** Given sets $A = \{1,3,4,5,6,7\}$ and $B = \{x \mid x \text{ is an odd positive integer and } x \leq 8\}$. 2. First, list set $B$ explicitly: $$ B = \{1,3,5,7\} $$ 3. Find $A \cup B$ (union): all elements in $A$ or $B$: $$ A \cup B = \{1,3,4,5,6,7\} \cup \{1,3,5,7\} = \{1,3,4,5,6,7\} $$ 4. Find $A \cap B$ (intersection): elements in both $A$ and $B$: $$ A \cap B = \{1,3,5,7\} $$ 5. Find $B - A$ (difference): elements in $B$ not in $A$: $$ B - A = \emptyset $$ 6. Find $A \oplus B$ (symmetric difference): elements in $A$ or $B$ but not both: $$ A \oplus B = (A - B) \cup (B - A) $$ 7. Calculate $A - B$: $$ A - B = \{4,6\} $$ 8. Since $B - A = \emptyset$, then: $$ A \oplus B = \{4,6\} $$ --- 1. **Problem 2c:** Find bitwise AND and XOR of bit strings $10100001$ and $10011000$. 2. Write bits aligned: $$\begin{array}{cccccccc} 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \end{array}$$ 3. Bitwise AND (1 if both bits are 1, else 0): $$ 10100001 \wedge 10011000 = 10000000 $$ 4. Bitwise XOR (1 if bits differ, else 0): $$ 10100001 \oplus 10011000 = 00111001 $$ **Final answers:** - 2a: The formula is neither tautology nor contradiction. - 2b: $A \cup B = \{1,3,4,5,6,7\}$, $A \cap B = \{1,3,5,7\}$, $B - A = \emptyset$, $A \oplus B = \{4,6\}$. - 2c: Bitwise AND = $10000000$, Bitwise XOR = $00111001$.